Problem: Recently Duff has been a soldier in the army. Malek is her commander.

Their country, Andarz Gu has n cities (numbered from 1 to n) and n - 1 bidirectional roads. Each road connects two different cities. There exist a unique path between any two cities.

There are also m people living in Andarz Gu (numbered from 1 to m). Each person has and ID number. ID number of i - th person is i and he/she lives in city number ci. Note that there may be more than one person in a city, also there may be no people living in the city.

<image>

Malek loves to order. That's why he asks Duff to answer to q queries. In each query, he gives her numbers v, u and a.

To answer a query:

Assume there are x people living in the cities lying on the path from city v to city u. Assume these people's IDs are p1, p2, ..., px in increasing order. 

If k = min(x, a), then Duff should tell Malek numbers k, p1, p2, ..., pk in this order. In the other words, Malek wants to know a minimums on that path (or less, if there are less than a people).

Duff is very busy at the moment, so she asked you to help her and answer the queries.

Input

The first line of input contains three integers, n, m and q (1 ≤ n, m, q ≤ 105).

The next n - 1 lines contain the roads. Each line contains two integers v and u, endpoints of a road (1 ≤ v, u ≤ n, v ≠ u).

Next line contains m integers c1, c2, ..., cm separated by spaces (1 ≤ ci ≤ n for each 1 ≤ i ≤ m).

Next q lines contain the queries. Each of them contains three integers, v, u and a (1 ≤ v, u ≤ n and 1 ≤ a ≤ 10).

Output

For each query, print numbers k, p1, p2, ..., pk separated by spaces in one line.

Examples

Input

5 4 5
1 3
1 2
1 4
4 5
2 1 4 3
4 5 6
1 5 2
5 5 10
2 3 3
5 3 1


Output

1 3
2 2 3
0
3 1 2 4
1 2

Note

Graph of Andarz Gu in the sample case is as follows (ID of people in each city are written next to them):

<image>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e5 + 10, mod = 1e9 + 7, INF = 0x3f3f3f3f;
int cnt[maxn * 22], ls[maxn * 22], rs[maxn * 22], rt[maxn], id;
void upd(int pre, int& now, int p, int l, int r) {
  now = ++id;
  cnt[now] = cnt[pre] + 1;
  ls[now] = ls[pre];
  rs[now] = rs[pre];
  if (l == r) return;
  int m = (l + r) >> 1;
  if (p <= m)
    upd(ls[pre], ls[now], p, l, m);
  else
    upd(rs[pre], rs[now], p, m + 1, r);
}
vector<int> v[maxn], e[maxn];
int fa[maxn][30], dep[maxn], m;
void dfs(int u, int f) {
  dep[u] = dep[f] + 1;
  fa[u][0] = f;
  rt[u] = rt[f];
  for (auto i : v[u]) upd(rt[u], rt[u], i, 1, m);
  for (int i = 1; i <= 22; ++i) fa[u][i] = fa[fa[u][i - 1]][i - 1];
  for (auto v : e[u]) {
    if (v == f) continue;
    dfs(v, u);
  }
}
int lca(int x, int y) {
  if (dep[x] < dep[y]) swap(x, y);
  for (int d = dep[x] - dep[y], i = 0; i <= 22; ++i)
    if ((d >> i) & 1) x = fa[x][i];
  if (x == y) return x;
  for (int i = 22; i >= 0; --i)
    if (fa[x][i] != fa[y][i]) x = fa[x][i], y = fa[y][i];
  return fa[x][0];
}
vector<int> ans;
void qry(int f, int lca, int u, int v, int k, int l, int r) {
  if (l == r) {
    if (cnt[u] + cnt[v] - cnt[lca] - cnt[f]) ans.push_back(l);
    return;
  }
  int m = (l + r) >> 1;
  int c = cnt[ls[u]] + cnt[ls[v]] - cnt[ls[lca]] - cnt[ls[f]];
  if (c < k) {
    qry(ls[f], ls[lca], ls[u], ls[v], c, l, m);
    qry(rs[f], rs[lca], rs[u], rs[v], k - c, m + 1, r);
  } else
    qry(ls[f], ls[lca], ls[u], ls[v], k, l, m);
}
int main() {
  int n, q;
  cin >> n >> m >> q;
  for (int i = 1; i < n; ++i) {
    int u, v;
    scanf("%d%d", &u, &v);
    e[u].push_back(v);
    e[v].push_back(u);
  }
  for (int i = 1; i <= m; ++i) {
    int x;
    scanf("%d", &x);
    v[x].push_back(i);
  }
  dfs(1, 0);
  for (int i = 1; i <= q; ++i) {
    int u, v, k;
    scanf("%d%d%d", &u, &v, &k);
    ans.clear();
    int l = lca(u, v);
    qry(rt[fa[l][0]], rt[l], rt[u], rt[v], k, 1, m);
    printf("%d", int(ans.size()));
    for (auto j : ans) printf(" %d", j);
    printf("\n");
  }
  return 0;
}